Import Question JSON

Current Question (ID: 10646)

Question:
$\text{Which of the following is the correct match?}$
Options:
  • 1. $\text{As}_2\text{O}_3 < \text{SiO}_2 < \text{P}_2\text{O}_3 < \text{SO}_2 \text{ - Basic strength.}$
  • 2. $\text{AsH}_3 < \text{PH}_3 < \text{NH}_3 \text{ - Enthalpy of vaporisation.}$
  • 3. $\text{S} < \text{O} < \text{Cl} < \text{F} \text{ - More negative electron gain enthalpy}$
  • 4. $\text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te} \text{ - Thermal stability}$ (Correct)
Solution:
$\text{HINT: Thermal stability of hydrides of group 16 decreases down the group.}$ $\text{Explanation:}$ $\text{(a) Acidic strength increases: As}_2\text{O}_3 < \text{SiO}_2 < \text{P}_2\text{O}_3 < \text{SO}_2$ $\text{(b) Correct order is enthalpy of vaporisation:}$ $\text{PH}_3 \text{ (14.6 kJ/mol)} < \text{AsH}_3 \text{ (16.6 kJ/mol)} < \text{NH}_3 \text{ (23.35 kJ/mol)}$ $\text{(c) O < S < F < Cl: Negative electron gain enthalpy order}$ $\text{Electron gain enthalpy becomes more and more negative from left to right in a period. As we move across a period from left to right the atomic size decreases and the nuclear charge increases. Both these factors tend to increase the attraction by the nucleus for the incoming electron. As we move down in a group of the periodic table the electron gain enthalpy should become less negative and fluorine is placed above chlorine yet has more negative electron gain enthalpy. Thus, the electron gain enthalpy of chlorine is more negative than that of fluorine. S has high negative electron gain enthalpy because in oxygen there is more inter-electronic repulsion than sulphur on gaining electrons.}$ $\text{(d) Thermal stability order: H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te}\text{. This is because M-H bond dissociation energy decreases down the group with the increase in the size of central atom.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}